Joe Student mixes 20.0 ml of a 3.00 M solution of silver nitrate with 15.0 ml of a 2.00 M solution of sodium phosphate. He collects and dries the precipitate formed, then weighs the precipitate and finds that it has a mass of 8.00 grams. The precipitate is Silver phosphate. Answer the following questions based on this information.

A. Write the balanced equation for this reaction.
3 AgNO3(aq)+ Na3PO4(aq)Ag3PO4(ppt.)+ 3 NaNO3(aq)

B. How many moles of each reactant were there?
(0.0200 L)(3.00 M) = 0.0600 mol AgNO3
(0.0150 L)(2.00 M) = 0.0300 mol Na3PO4

C. What is the limiting reagent?
0.0300 mol Na3PO4(3 mol AgNO3/ 1 mol Na3PO4) = 0.0900 mol AgNO3neededonly have 0.0600 mol AgNO3so it is the limiting reagent

D. How many moles of excess reagent were left unreacted?
0.0600 mol L. R. (1 mol E. R. / 3 mol L. R.) = 0.0200 mol
E. R. needed0.0300 mol - 0.0200 mol = 0.0100 mol xcs

E. What is the precipitate?
Ag3PO4(ppt.)

F. How many moles of precipitate should have been formed theoretically?

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

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